byepgp.inflibnet.ac.in/epgpdata/uploads/epgp_content/S...Numerical Analysis by Dr. Anita Pal Assistant Professor Department of Mathematics National Institute of Technology Durgapur

Numerical Analysis

by

Dr. Anita Pal

Assistant Professor

Department of Mathematics

National Institute of Technology Durgapur

Durgapur-713209

email: [email protected]

.

Chapter 1

Numerical Errors

Module No. 3

Operators in Numerical Analysis

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

Lot of operators are used in numerical analysis/computation. Some of the frequently

used operators, viz. forward difference (∆), backward difference (∇), central difference

(δ), shift (E) and mean (µ) are discussed in this module.

Let the function y = f(x) be defined on the closed interval [a, b] and let x0, x1, . . . , xn

be the n values of x. Assumed that these values are equidistance, i.e. xi = x0 + ih,

i = 0, 1, 2, . . . , n; h is a suitable real number called the difference of the interval or

spacing. When x = xi, the value of y is denoted by yi and is defined by yi = f(xi). The

values of x and y are called arguments and entries respectively.

3.1 Finite difference operators

Different types of finite difference operators are defined, among them forward dif-

ference, backward difference and central difference operators are widely used. In this

section, these operators are discussed.

3.1.1 Forward difference operator

The forward difference is denoted by ∆ and is defined by

∆f(x) = f(x+ h)− f(x). (3.1)

When x = xi then from above equation

∆f(xi) = f(xi + h)− f(xi), i.e. ∆yi = yi+1 − yi, i = 0, 1, 2, . . . , n− 1. (3.2)

In particular, ∆y0 = y1− y0,∆y1 = y2− y1, . . . ,∆yn−1 = yn− yn−1. These are called

first order differences.

The differences of the first order differences are called second order differences. The

second order differences are denoted by ∆2y0,∆2y1, . . ..

Two second order differences are

∆2y0 = ∆y1 −∆y0 = (y2 − y1)− (y1 − y0) = y2 − 2y1 + y0

∆2y1 = ∆y2 −∆y1 = (y3 − y2)− (y2 − y1) = y3 − 2y2 + y1.

The third order differences are also defined in similar manner, i.e.

∆3y0 = ∆2y1 −∆2y0 = (y3 − 2y2 + y1)− (y2 − 2y1 + y0) = y3 − 3y2 + 3y1 − y0∆3y1 = y4 − 3y3 + 3y2 − y1.

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. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .Operators in Numerical Analysis

Similarly, higher order differences can be defined.

In general,

∆n+1f(x) = ∆[∆nf(x)], i.e. ∆n+1yi = ∆[∆nyi], n = 0, 1, 2, . . . . (3.3)

Again, ∆n+1f(x) = ∆n[f(x+ h)− f(x)] = ∆nf(x+ h)−∆nf(x)

and

∆n+1yi = ∆nyi+1 −∆nyi, n = 0, 1, 2, . . . . (3.4)

It must be remembered that ∆0 ≡ identity operator, i.e. ∆0f(x) = f(x) and ∆1 ≡ ∆.

All the forward differences can be represented in a tabular form, called the forward

difference or diagonal difference table.

Let x0, x1, . . . , x4 be four arguments. All the forwarded differences of these arguments

are shown in Table 3.1.

x y ∆ ∆2 ∆3 ∆4

x0 y0

∆y0

x1 y1 ∆2y0

∆y1 ∆3y0

x2 y2 ∆2y1 ∆4y0

∆y2 ∆3y1

x3 y3 ∆2y2

∆y3

x4 y4

Table 3.1: Forward difference table.

3.1.2 Error propagation in a difference table

If any entry of the difference table is erroneous, then this error spread over the table

in convex manner.

The propagation of error in a difference table is illustrated in Table 3.2. Let us

assumed that y3 be erroneous and the amount of the error be ε.

Following observations are noted from Table 3.2.

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x y ∆y ∆2y ∆3y ∆4y ∆5y

x0 y0

∆y0

x1 y1 ∆2y0

∆y1 ∆3y0 + ε

x2 y2 ∆2y1 + ε ∆4y0 − 4ε

∆y2 + ε ∆3y1 − 3ε ∆5y0 + 10ε

x3 y3 + ε ∆2y2 − 2ε ∆4y1 + 6ε

∆y3 − ε ∆3y2 + 3ε ∆5y1 − 10ε

x4 y4 ∆2y3 + ε ∆4y2 − 4ε

∆y4 ∆3y3 − εx5 y5 ∆2y4

∆y5

x6 y6

Table 3.2: Error propagation in a finite difference table.

(i) The error increases with the order of the differences.

(ii) The error is maximum (in magnitude) along the horizontal line through the erro-

neous tabulated value.

(iii) In the kth difference column, the coefficients of errors are the binomial coefficients

in the expansion of (1 − x)k. In particular, the errors in the second difference

column are ε,−2ε, ε, in the third difference column these are ε,−3ε, 3ε,−ε, and

so on.

(iv) The algebraic sum of errors in any complete column is zero.

If there is any error in a single entry of the table, then we can detect and correct

it from the difference table. The position of the error in an entry can be identified by

performing the following steps.

(i) If at any stage, the differences do not follow a smooth pattern, then there is an

error.

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(ii) If the differences of some order (it is generally happens in higher order) becomes

alternating in sign then the middle entry contains an error.

Properties

Some common properties of forward difference operator are presented below:

(i) ∆c = 0, where c is a constant.

(ii) ∆[f1(x) + f2(x) + · · ·+ fn(x)]

= ∆f1(x) + ∆f2(x) + · · ·+ ∆fn(x).

(iii) ∆[cf(x)] = c∆f(x).

Combining properties (ii) and (iii), one can generalise the property (ii) as

(iv) ∆[c1f1(x) + c2f2(x) + · · ·+ cnfn(x)]

= c1∆f1(x) + c2∆f2(x) + · · ·+ cn∆fn(x).

(v) ∆m∆nf(x) = ∆m+nf(x) = ∆n∆mf(x) = ∆k∆m+n−kf(x),

k = 0, 1, 2, . . . ,m or n.

(vi) ∆[cx] = cx+h − cx = cx(ch − 1), for some constant c.

(vii) ∆[xCr] = xCr−1, where r is fixed and h = 1.

∆[xCr] = x+1Cr − xCr = xCr−1 as h = 1.

Example 3.1

∆[f(x)g(x)] = f(x+ h)g(x+ h)− f(x)g(x)

= f(x+ h)g(x+ h)− f(x+ h)g(x) + f(x+ h)g(x)− f(x)g(x)

= f(x+ h)[g(x+ h)− g(x)] + g(x)[f(x+ h)− f(x)]

= f(x+ h)∆g(x) + g(x)∆f(x).

Also, it can be shown that

∆[f(x)g(x)] = f(x)∆g(x) + g(x+ h)∆f(x)

= f(x)∆g(x) + g(x)∆f(x) + ∆f(x)∆g(x).

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Example 3.2 ∆

[f(x)

g(x)

]=g(x)∆f(x)− f(x)∆g(x)

g(x+ h)g(x), g(x) 6= 0.

∆

[f(x)

g(x)

]=f(x+ h)

g(x+ h)− f(x)

g(x)

=f(x+ h)g(x)− g(x+ h)f(x)

g(x+ h)g(x)

=g(x)[f(x+ h)− f(x)]− f(x)[g(x+ h)− g(x)]

g(x+ h)g(x)

=g(x)∆f(x)− f(x)∆g(x)

g(x+ h)g(x).

In particular, when the numerator is 1, then

∆

[1

f(x)

]= − ∆f(x)

f(x+ h)f(x).

3.1.3 Backward difference operator

The symbol ∇ is used to represent backward difference operator. The backward

difference operator is defined as

∇f(x) = f(x)− f(x− h). (3.5)

When x = xi, the above relation reduces to

∇yi = yi − yi−1, i = n, n− 1, . . . , 1. (3.6)

In particular,

∇y1 = y1 − y0,∇y2 = y2 − y1, . . . ,∇yn = yn − yn−1. (3.7)

These are called the first order backward differences. The second order differences

are denoted by ∇2y2,∇2y3, . . . ,∇2yn. First two second order backward differences are

∇2y2 = ∇(∇y2) = ∇(y2− y1) = ∇y2−∇y1 = (y2− y1)− (y1− y0) = y2− 2y1 + y0, and

∇2y3 = y3 − 2y2 + y1,∇2y4 = y4 − 2y3 + y2.

The other second order differences can be obtained in similar manner.

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In general,

∇kyi = ∇k−1yi −∇k−1yi−1, i = n, n− 1, . . . , k, (3.8)

where ∇0yi = yi,∇1yi = ∇yi.Like forward differences, these backward differences can be written in a tabular form,

called backward difference or horizontal difference table.

All backward difference table for the arguments x0, x1, . . . , x4 are shown in Table 3.3.

x y ∇ ∇2 ∇3 ∇4

x0 y0

x1 y1 ∇y1x2 y2 ∇y2 ∇2y2

x3 y3 ∇y3 ∇2y3 ∇3y3

x4 y4 ∇y4 ∇2y4 ∇3y4 ∇4y4

Table 3.3: Backward difference table.

It is observed from the forward and backward difference tables that for a given table

of values both the tables are same. Practically, there are no differences among the values

of the tables, but, theoretically they have separate significant.

3.1.4 Central difference operator

There is another kind of finite difference operator known as central difference operator.

This operator is denoted by δ and is defined by

δf(x) = f(x+ h/2)− f(x− h/2). (3.9)

When x = xi, then the first order central difference, in terms of ordinates is

δyi = yi+1/2 − yi−1/2 (3.10)

where yi+1/2 = f(xi + h/2) and yi−1/2 = f(xi − h/2).

In particular, δy1/2 = y1 − y0, δy3/2 = y2 − y1, . . . , δyn−1/2 = yn − yn−1.The second order central differences are

δ2yi = δyi+1/2 − δyi−1/2 = (yi+1 − yi)− (yi − yi−1) = yi+1 − 2yi + yi−1.

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In general,

δnyi = δn−1yi+1/2 − δn−1yi−1/2. (3.11)

All central differences for the five arguments x0, x1, . . . , x4 is shown in

Table 3.4.

x y δ δ2 δ3 δ4

x0 y0

δy1/2

x1 y1 δ2y1

δy3/2 δ3y3/2

x2 y2 δ2y2 δ4y2

δy5/2 δ3y5/2

x3 y3 δ2y3

δy7/2

x4 y4

Table 3.4: Central difference table.

It may be observed that all odd (even) order differences have fraction suffices (integral

suffices).

3.1.5 Shift, average and differential operators

Shift operator, E:

The shift operator is denoted by E and is defined by

Ef(x) = f(x+ h). (3.12)

In terms of y, the above formula becomes

Eyi = yi+1. (3.13)

Note that shift operator increases subscript of y by one. When the shift operator is

applied twice on the function f(x), then the subscript of y is increased by 2.

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That is,

E2f(x) = E[Ef(x)] = E[f(x+ h)] = f(x+ 2h). (3.14)

In general,

Enf(x) = f(x+ nh) or Enyi = yi+nh. (3.15)

The inverse shift operator can also be find in similar manner. It is denoted by E−1

and is defined by

E−1f(x) = f(x− h). (3.16)

Similarly, second and higher order inverse operators are defined as follows:

E−2f(x) = f(x− 2h) and E−nf(x) = f(x− nh). (3.17)

The general definition of shift operator is

Erf(x) = f(x+ rh), (3.18)

where r is positive as well as negative rational numbers.

Properties

Few common properties of E operator are given below:

(i) Ec = c, where c is a constant.

(ii) E{cf(x)} = cEf(x).

(iii) E{c1f1(x) + c2f2(x) + · · ·+ cnfn(x)]

= c1Ef1(x) + c2Ef2(x) + · · ·+ cnEfn(x).

(iv) EmEnf(x) = EnEmf(x) = Em+nf(x).

(v) EnE−nf(x) = f(x).

In particular, EE−1 ≡ I, I is the identity operator and it is some times denoted

by 1.

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(vi) (En)mf(x) = Emnf(x).

(vii) E

{f(x)

g(x)

}=Ef(x)

Eg(x).

(viii) E{f(x) g(x)} = Ef(x) Eg(x).

(ix) E∆f(x) = ∆Ef(x).

(x) ∆mf(x) = ∇mEmf(x) = Em∇mf(x)

and ∇mf(x) = ∆mE−mf(x) = E−m∆mf(x).

Average operator, µ:

The average operator is denoted by µ and is defined by

µf(x) =1

2

[f(x+ h/2) + f(x− h/2)

]In terms of y, the above definition becomes

µyi =1

2

[yi+1/2 + yi−1/2

].

Here the average of the values of f(x) at two points (x+h/2) and f(x−h/2) is taken

as the value of µf(x).

Differential operator, D:

The differential operator is well known from differential calculus and it is denoted by

D. This operator gives the derivative. That is,

Df(x) =d

dxf(x) = f ′(x) (3.19)

D2f(x) =d2

dx2f(x) = f ′′(x) (3.20)

· · · · · · · · · · · · · · · · · · · · ·

Dnf(x) =dn

dxnf(x) = fn(x). (3.21)

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3.1.6 Factorial notation

The factorial notation is a very useful notation in calculus of finite difference. Using

this notation one can find all order differences by the rules used in differential calculus.

It is also very useful and simple notation to find anti-differences. The nth factorial of x

is denoted by x(n) and is defined by

x(n) = x(x− h)(x− 2h) · · · (x− n− 1h), (3.22)

where, each factor is decreased from the earlier by h; and x(0) = 1.

Similarly, the nth negative factorial of x is defined by

x(−n) =1

x(x+ h)(x+ 2h) · · · (x+ n− 1h). (3.23)

A very interesting and obvious relation is x(n).x(−n) 6= 1.

Following results show the similarity of factorial notation and differential operator.

Property 3.1 ∆x(n) = nhx(n−1).

Proof.

∆x(n) = (x+ h)(x+ h− h)(x+ h− 2h) · · · (x+ h− n− 1h)

−x(x− h)(x− 2h) · · · (x− n− 1h)

= x(x− h)(x− 2h) · · · (x− n− 2h)[x+ h− {x− (n− 1)h}]

= nhx(n−1).

Note that this property is analogous to the differential formula D(xn) = nxn−1 when

h = 1.

The above formula can also be used to find anti-difference (like integration in integral

calculus), as

∆−1x(n−1) =1

nhx(n). (3.24)

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3.2 Relations among operators

Lot of useful and interesting results can be derived among the operators discussed

above. First of all, we determine the relation between forward and backward difference

operators.

∆yi = yi+1 − yi = ∇yi+1 = δyi+1/2

∆2yi = yi+2 − 2yi+1 + yi = ∇2yi+2 = δ2yi+1

etc.

In general,∆nyi = ∇nyi+n, i = 0, 1, 2, . . . . (3.25)

There is a good relation between E and ∆ operators.∆f(x) = f(x+ h)− f(x) = Ef(x)− f(x) = (E − 1)f(x).

From this relation one can conclude that the operators ∆ and E − 1 are equivalent.

That is,

∆ ≡ E − 1 or E ≡ ∆ + 1. (3.26)

The relation between ∇ and E operators is derived below:∇f(x) = f(x)− f(x− h) = f(x)− E−1f(x) = (1− E−1)f(x).

That is,∇ ≡ 1− E−1. (3.27)

The expression for higher order forward differences in terms of function values can

be derived as per following way:

∆3yi = (E − 1)3yi = (E3 − 3E2 + 3E − 1)yi = y3 − 3y2 + 3y1 − y0.

The relation between the operators δ and E is given below:

δf(x) = f(x+ h/2)− f(x− h/2) = E1/2f(x)− E−1/2f(x) = (E1/2 − E−1/2)f(x).

That is,

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δ ≡ E1/2 − E−1/2. (3.28)

The average operator µ is expressed in terms of E and δ as follows:

µf(x) =1

2

[f(x+ h/2) + f(x− h/2)

]=

1

2

[E1/2f(x) + E−1/2f(x)

]=

1

2(E1/2 + E−1/2)f(x).

Thus,µ ≡ 1

2

[E1/2 + E−1/2

]. (3.29)

µ2f(x) =1

4

[E1/2 + E−1/2

]2f(x)

=1

4

[(E1/2 − E−1/2)2 + 4

]f(x) =

1

4

[δ2 + 4

]f(x).

Hence,

µ ≡√

1 +1

4δ2. (3.30)

Every operator defined earlier can be expressed in terms of other operator(s). Few

more relations among the operators ∆,∇, E and δ are deduced in the following.

∇Ef(x) = ∇f(x+ h) = f(x+ h)− f(x) = ∆f(x).

Also,

δE1/2f(x) = δf(x+ h/2) = f(x+ h)− f(x) = ∆f(x).

Thus,

∆ ≡ ∇E ≡ δE1/2. (3.31)

There is a very nice relation among the operators E and D, deduced below.

Ef(x) = f(x+ h) = f(x) + hf ′(x) +h2

2!f ′′(x) +

h3

3!f ′′′(x) + · · ·

[by Taylor’s series]

= f(x) + hDf(x) +h2

2!D2f(x) +

h3

3!D3f(x) + · · ·

=

[1 + hD +

h2

2!D2 +

h3

3!D3 + · · ·

]f(x)

= ehDf(x).

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Hence,E ≡ ehD. (3.32)

This result can also be written ashD ≡ logE. (3.33)

The relation between the operators D and δ is deduced below:

δf(x) = [E1/2 − E−1/2]f(x) =[ehD/2 − e−hD/2

]f(x)

= 2 sinh(hD

2

)f(x).

Thus,

δ ≡ 2 sinh(hD

2

). Similarly, µ ≡ cosh

(hD2

). (3.34)

Again,

µδ ≡ 2 cosh(hD

2

)sinh

(hD2

)= sinh(hD). (3.35)

This relation gives the inverse result,

hD ≡ sinh−1(µδ). (3.36)

From the relation (3.33) and using the relations E ≡ 1 + ∆ and E−1 ≡ 1 − ∇ we

obtained,

hD ≡ logE ≡ log(1 + ∆) ≡ − log(1−∇) ≡ sinh−1(µδ). (3.37)

Some of the operators are commutative with other operators. For example, µ and E

are commutative, as

µEf(x) = µf(x+ h) =1

2

[f(x+ 3h/2) + f(x+ h/2)

],

and

Eµf(x) = E[1

2

{f(x+ h/2) + f(x− h/2)

}]=

1

2

[f(x+ 3h/2) + f(x+ h/2)

].

Hence,µE ≡ Eµ. (3.38)

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Example 3.3 Prove the following relations.

(i) (1 + ∆)(1−∇) ≡ 1

(ii) µ ≡ cosh(hD

2

)(iii) µδ ≡ ∆ +∇

2(iv) ∆∇ ≡ ∇∆ ≡ δ2

(v) µδ ≡ ∆E−1

2+

∆

2

(vi) E1/2 ≡ µ+δ

2

(vvi) 1 + δ2µ2 ≡(

1 +δ2

2

)2

(viii) ∆ ≡ δ2

2+ δ

√1 +

δ2

4.

Solution. (i) (1 + ∆)(1−∇)f(x) = (1 + ∆)[f(x)− f(x) + f(x− h)]

= (1 + ∆)f(x− h) = f(x− h) + f(x)− f(x− h)

= f(x).

Therefore,

(1 + ∆)(1−∇) ≡ 1. (3.39)

(ii)

µf(x) =1

2[E1/2 + E−1/2]f(x) =

1

2

[ehD/2 + e−hD/2

]f(x)

= cosh(hD

2

)f(x).

(iii) [∆ +∇

2

]f(x) =

1

2[∆f(x) +∇f(x)]

=1

2[f(x+ h)− f(x) + f(x)− f(x− h)]

=1

2[f(x+ h)− f(x− h)] =

1

2[E − E−1]f(x)

= µδf(x) (as in previous case).

Thus,

µδ ≡ ∆ +∇2

. (3.40)

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(iv) ∆∇f(x) = ∆[f(x)− f(x− h)] = f(x+ h)− 2f(x) + f(x− h).

Again,

∇∆f(x) = f(x+ h)− 2f(x) + f(x− h) = (E − 2 + E−1)f(x)

= (E1/2 − E−1/2)2f(x) = δ2f(x).

Hence,∆∇ ≡ ∇∆ ≡ (E1/2 − E−1/2)2 ≡ δ2. (3.41)

(v) [∆E−1

2+

∆

2

]f(x) =

1

2[∆f(x− h) + ∆f(x)]

=1

2[f(x)− f(x− h) + f(x+ h)− f(x)]

=1

2[f(x+ h)− f(x− h)] =

1

2[E − E−1]f(x)

=1

2(E1/2 + E−1/2)(E1/2 − E−1/2)f(x)

= µδf(x).

Hence

∆E−1

2+

∆

2≡ µδ. (3.42)

(vi)

(µ+

δ

2

)f(x) =

{1

2[E1/2 + E−1/2] +

1

2[E1/2 − E−1/2]

}f(x) = E1/2f(x).

Thus

E1/2 ≡ µ+δ

2. (3.43)

(vii) δµf(x) = 12(E1/2 + E−1/2)(E1/2 − E−1/2)f(x) = 1

2 [E − E−1]f(x).

Therefore,

(1 + δ2µ2)f(x) =

[1 +

1

4(E − E−1)2

]f(x)

=

[1 +

1

4(E2 − 2 + E−2)

]f(x) =

1

4(E + E−1)2f(x)

=

[1 +

1

2(E1/2 − E−1/2)2

]2f(x) =

[1 +

δ2

2

]2f(x).

15


Hence

1 + δ2µ2 ≡(

1 +δ2

2

)2

. (3.44)

(viii) [δ2

2+ δ

√1 +

δ2

4

]f(x)

=1

2(E1/2 − E−1/2)2f(x) +

[(E1/2 − E−1/2)

√1 +

1

4(E1/2 − E−1/2)2

]f(x)

=1

2[E + E−1 − 2]f(x) +

1

2(E1/2 − E−1/2)(E1/2 + E−1/2)f(x)

=1

2[E + E−1 − 2]f(x) +

1

2(E − E−1)f(x)

= (E − 1)f(x).

Hence, δ2

2+ δ

√1 +

δ2

4≡ E − 1 ≡ ∆. (3.45)

In Table 3.5, it is shown that any operator can be expressed with the help of another

operator.

E ∆ ∇ δ hD

E E ∆ + 1 (1−∇)−1 1+δ2

2+ δ

√1+δ2

4ehD

∆ E − 1 ∆ (1−∇)−1 − 1δ2

2+ δ

√1 +

δ2

4ehD − 1

∇ 1− E−1 1− (1 + ∆)−1 ∇ −δ2

2+ δ

√1 +

δ2

41− e−hD

δ E1/2−E−1/2 ∆(1 + ∆)−1/2 ∇(1−∇)−1/2 δ 2 sinh(hD/2)

µE1/2+E−1/2

2(1 + ∆/2) (1−∇/2)(1−∇)−1/2 1 +

δ2

4cosh(hD/2)

×(1 + ∆)−1/2

hD logE log(1 + ∆) − log(1−∇) 2 sinh−1(δ/2) hD

Table 3.5: Relationship between the operators.

From earlier discussion we noticed that there is an approximate equality between ∆

operator and derivative. These relations are presented below.

16

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

By the definition of derivative,

f ′(x) = limh→0

f(x+ h)− f(x)

h= lim

h→0

∆f(x)

h.

Thus, ∆f(x) ' hf ′(x) = hDf(x).

Again,

f ′′(x) = limh→0

f ′(x+ h)− f ′(x)

h

' limh→0

∆f(x+ h)

h− ∆f(x)

hh

= limh→0

∆f(x+ h)−∆f(x)

h2= lim

h→0

∆2f(x)

h2.

Hence, ∆2f(x) ' h2f ′′(x) = h2D2f(x).

In general, ∆nf(x) ' hnfn(x) = hnDnf(x). That is, for small values of h, the

operators ∆ and hD are almost equal.

3.3 Polynomial using factorial notation

According to the definition of factorial notation, one can write

x(0) = 1

x(1) = x

x(2) = x(x− h)

x(3) = x(x− h)(x− 2h)

x(4) = x(x− h)(x− 2h)(x− 3h)

(3.46)

and so on.

From these equations it is obvious that the base terms (x, x2, x3, . . .) of a polynomial

can be expressed in terms of factorial notations x(1), x(2), x(3), . . ., as shown below.

1 = x(0)

x = x(1)

x2 = x(2) + hx(1)

x3 = x(3) + 3hx(2) + h2x(1)

x4 = x(4) + 6hx(3) + 7h2x(2) + h3x(1)

(3.47)

17


and so on.

Note that the degree of xk (for any k = 1, 2, 3, . . .) remains unchanged while expressed

it in factorial notation. This observation leads to the following lemma.

Lemma 3.1 Any polynomial f(x) in x can be expressed in factorial notation with same

degree.

Since all the base terms of a polynomial are expressed in terms of factorial notation,

every polynomial can be written with the help of factorial notation. Once a polynomial

is expressed in a factorial notation, then its differences can be determined by using the

formula like differential calculus.

Example 3.4 Express f(x) = 10x4 − 41x3 + 4x2 + 3x + 7 in factorial notation and

find its first and second differences.

Solution. For simplicity, we assume that h = 1.

Now by (3.47), x = x(1), x2 = x(2) + x(1), x3 = x(3) + 3x(2) + x(1),

x4 = x(4) + 6x(3) + 7x(2) + x(1).

Substituting these values to the function f(x) and we obtained

f(x) = 10[x(4) + 6x(3) + 7x(2) + x(1)

]− 41

[x(3) + 3x(2) + x(1)

]+ 4[x(2) + x(1)

]+ 3x(1) + 7

= 10x(4) + 19x(3) − 49x(2) − 24x(1) + 7.

Now, the relation ∆x(n) = nx(n−1) (Property 3.1) is used to find the first and second

order differences. Therefore,

∆f(x) = 10.4x(3) + 19.3x(2) − 49.2x(1) − 24.1x(0) = 40x(3) + 57x(2) − 98x(1) − 24

= 40x(x− 1)(x− 2) + 57x(x− 1)− 98x− 24 = 40x3 − 63x2 − 75x− 24

and ∆2f(x) = 120x(2) + 114x(1) − 98 = 120x(x− 1) + 114x− 98 = 120x2 − 6x− 98.

The above process to convert a polynomial in a factorial notation is a very labourious

task when the degree of the polynomial is large. There is a systematic method, similar to

Maclaurin’s formula in differential calculus, is used to convert a polynomial in factorial

notation. This technique is also useful for a function which satisfies the Maclaurin’s

theorem for infinite series.

Let f(x) be a polynomial in x of degree n. We assumed that in factorial notation

f(x) is of the following form

f(x) = a0 + a1x(1) + a2x

(2) + · · ·+ anx(n), (3.48)

18

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

where ai’s are unknown constants to be determined and an 6= 0.

Now, we determine the different differences of (3.48) as follows.

∆f(x) = a1 + 2a2x(1) + 3a3x

(2) + · · ·+ nanx(n−1)

∆2f(x) = 2.1a2 + 3.2a3x(1) + · · ·+ n(n− 1)anx

(n−2)

∆3f(x) = 3.2.1a3 + 4.3.2.x(1) + · · ·+ n(n− 1)(n− 2)anx(n−3)

· · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · ·

∆nf(x) = n(n− 1)(n− 2) · · · 3 · 2 · 1an = n!an.

Substituting x = 0 to the above relations and we obtained

a0 = f(0), ∆f(0) = a1,

∆2f(0) = 2.1.a2 or, a2 =∆2f(0)

2!

∆3f(0) = 3.2.1.a3 or, a3 =∆3f(0)

3!· · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · ·

∆nf(0) = n!an or, an =∆nf(0)

n!.

Using these results equation (3.48) transferred to

f(x) = f(0) + ∆f(0)x(1) +∆2f(0)

2!x(2) +

∆3f(0)

3!x(3) + · · ·+ ∆nf(0)

n!x(n).

(3.49)

Observed that this formula is similar to Maclaurin’s formula of differential calculus.

This formula can also be used to expand a function in terms of factorial notation. To

expand a function in terms of factorial notation different forward differences are needed

at x = 0. These differences can be determined using the forward difference table and

the entire method is explained with the help of the following example.

Example 3.5 Express f(x) = 15x4 − 3x3 − 6x2 + 11 in factorial notation.

Solution. Let h = 1. For the given function, f(0) = 11, f(1) = 17, f(2) = 203, f(3) =

1091, f(4) = 3563.

19


x f(x) ∆f(x) ∆2f(x) ∆3f(x) ∆4f(x)

0 11

6

1 17 180

186 522

2 203 702 360

888 882

3 1091 1584

2472

4 3563

Thus by formula (3.49)f(x) = f(0) + ∆f(0)x(1) +

∆2f(0)

2!x(2) +

∆3f(0)

3!x(3) +

∆4f(0)

4!x(4)

= 15x(4) + 87x(3) + 90x(2) + 6x(1) + 11.

There is another method to find the coefficients of a polynomial in factorial notation,

presented below.

Example 3.6 Find f(x), if ∆f(x) = x4 − 10x3 + 11x2 + 5x+ 3.

Solution. The synthetic division is used to express ∆f(x) in factorial notation.

1 1 −10 11 5 3

1 −9 2

2 1 −9 2 7

2 −14

3 1 −7 −12

3

4 1 −4

1

Therefore, ∆f(x) = x(4) − 4x(3) − 12x(2) + 7x(1) + 3.

Hence,

f(x) =1

5x(5) − 4

4x(4) − 12

3x(3) +

7

2x(2) + 3x(1) + c, [using Property 1]

=1

5x(x− 1)(x− 2)(x− 3)(x− 4)− x(x− 1)(x− 2)(x− 3)

−4x(x− 1)(x− 2) +7

2x(x− 1) + 3x+ c, where c is arbitrary constant.

20

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

3.4 Difference of a polynomial

Let f(x) = a0xn + a1x

n−1 + · · ·+ an−1x+ an be a polynomial in x of degree n, where

ai’s are the given coefficients.

Suppose, f(x) = b0x(n) + b1x

(n−1) + b2x(n−2) + · · · + bn−1x

(1) + bn be the same

polynomial in terms of factorial notation. The coefficients bi’s can be determined by

using any method discussed earlier.

Now,

∆f(x) = b0nhx(n−1) + b1h(n− 1)x(n−2) + b2h(n− 2)x(n−3) + · · ·+ bn−1h.

Clearly this is a polynomial of degree n− 1.

Similarly,

∆2f(x) = b0n(n− 1)h2x(n−2) + b1(n− 1)(n− 2)h2x(n−3) + · · ·+ bn−2h2,

∆3f(x) = b0n(n− 1)(n− 2)h3x(n−3) + b1(n− 1)(n− 2)(n− 3)h3x(n−4)

+ · · ·+ bn−3h3.

In this way, ∆kf(x) = b0n(n− 1)(n− 2) · · · (n− k + 1)hkx(n−k).

Thus finally,

∆kf(x), k < n is a polynomial of degree n− k,

∆nf(x) = b0n!hn = n!hna0 is constant, and

∆kf(x) = 0, if k > n.

In particular, ∆n+1f(x) = 0.

Example 3.7 Let ui(x) = (x − x0)(x − x1) · · · (x − xi), where xi = x0 + ih, i =

0, 1, 2, . . . , n;h > 0. Prove that

∆kui(x) = (i+ 1)i(i− 1) · · · (i− k + 2)hk(x− x0)(x− x1) · · · (x− xi−k).

Solution. Let ui(x) = (x− x0)(x− x1) · · · (x− xi) be denoted by (x− x0)(i+1).

21


Therefore,

∆ui(x) = (x+ h− x0)(x+ h− x1) · · · (x+ h− xi)− (x− x0) · · · (x− xi)

= (x+ h− x0)(x− x0)(x− x1) · · · (x− xi−1)

−(x− x0)(x− x1) · · · (x− xi)

= (x− x0)(x− x1) · · · (x− xi−1)[(x+ h− x0)− (x− xi)]

= (x− x0)(x− x1) · · · (x− xi−1)(h+ xi − x0)

= (x− x0)(x− x1) · · · (x− xi−1)(i+ 1)h [since xi = x0 + ih]

= (i+ 1)h(x− x0)(i).

By similar way,

∆2ui(x) = (i+ 1)h[(x+ h− x0)(x+ h− x1) · · · (x+ h− xi−1)

−(x− x0)(x− x1) · · · (x− xi−1)]

= (i+ 1)h(x− x0)(x− x1) · · · (x− xi−2)[(x+ h− x0)− (x− xi−1)]

= (i+ 1)h(x− x0)(i−1)ih

= (i+ 1)ih2(x− x0)(i−1).

Also, ∆3ui(x) = (i+ 1)i(i− 1)h3(x− x0)(i−2).Hence, in this way

∆kui(x) = (i+ 1)i(i− 1) · · · (i− k − 2)hk(x− x0)(i−k−1)

= (i+ 1)i(i− 1) · · · (i− k + 2)hk(x− x0)(x− x1) · · · (x− xi−k).

22

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byepgp.inflibnet.ac.in/epgpdata/uploads/epgp_content/S...Numerical Analysis by Dr. Anita Pal Assistant Professor Department of Mathematics National Institute of Technology Durgapur